Question

Use the data given in the following table to calculate the molar mass of naturally occurring argon having isotopes:

Isotope	Isotopic molar mass	Abundance
$\,{}^{36}Ar$	$35.96755\, g\, mol^{-1}$	$0.337\%$
$\,{}^{38}Ar$	$37.96272 \,g\, mol^{-1}$	$0.063\%$
$\,{}^{40}Ar$	$ 39.9624\, g\, mol^{-1}$	$99.600\%$

• A. $89.94 \,g \,mol^{-1}$
• B. $39.95 \,g \,mol^{-1}$
• C. $94.39\, g\, mol^{-1}$
• D. $49.39 \,g \,mol^{-1}$

Answer 1

Answer: (B)

Molar mass of naturally occurring argon
$= \frac{35.96755 \times 0.337 + 37.96272 \times 0.063 + 39.9624 \times 99.600}{0.337 + 0.063 + 99.600}$
$ = \frac{12.12 + 2.39 + 3980.255}{100} $
$ = \frac{3994.765}{100} $
$= 39.95\,g\,mol^{-1}$